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Sonobe Origami for Enriching Understanding of Geometric Concepts in Three Dimensions

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Sonobe Origami for Enriching Understanding of Geometric Concepts in Three Dimensions Sonobe Origami for enriching understanding of geometric concepts in three dimensions DONNA A. DIETZ American University Washington, D.C. Donna Dietz, American University Sonobe Origami for enriching understanding of geometric concepts in three dimensions St. Mary's College of Maryland November 7, 2015. MAA Section Meeting 3:15 pm -3:35 pm Goodpaster Hall Room 109 "Programs that take advantage of paper folding to teach mathematics are thriving in many parts of the world," according to the organizers of the MAA origami-themed Contributed Paper Session to be held at the JMM in January 2016. But, K-12 should not be having all the fun! In this talk, I will show some ways of stimulating student engagement using sonobe origami. These activities can be used in General Education mathematics "appreciation" courses or for Non-Euclidean Geometry for mathematics majors. Specifically, the goals are enriching student understanding of surface curvatures and helping them understand the duality of the regular polyhedra using these folded paper objects. Things to chat about... ● Construction overview and classroom hints ● Duality of regular polyhedra (aka Platonic Solids) ● Euler's χ for topological classification ● Angle deficit for topological classification ● Gauss-Bonnet Theorem ● Bonus Puzzle! Construction Basics ● tinyurl.com/sonobe ● I also have videos on YouTube ● for Gen Ed course- one class period ● students spend about an hour outside of class ● 30 pieces of paper per student ● chirality compatibility is a construction issue ● assembly is a learning experience for students The Paper I Use in Class: Pros and Cons ● No glue ● Only time for one large project rather ● No scissors than many small ones ● Straightedge and Compass is ● Object is more equivalent to folding complex than regular polyhedra ● Pride in crafting Make 30 of those Ok, Great! Now, what is this sonobe project good for? Duality of Regular Polyhedra mathforum.org Cube with Octahedron behance.net Icosahedron with Dodecahedron behance.net Tetrahedron is self dual math.brown.edu mathcraft.wonderhowto.com How much paper is that!? To build the basic object, consider the underlying icosahedron (20 sides) and count pyramids. Each side has a pyramid attached which uses 3 units. But each unit is used in two pyramids. 20*3/2 = 30 Other objects ● Tetrahedron: 4*3/2 = 6 units, but it's a cube! ● Octahedron: 8*3/2 = 12 units ● Icosahedron: 20*3/2 = 30 units ● Massive ball: (12x5)*3/2 = 90 units All of these objects are based off of an underlying triangular mesh frame. Euler's χ for topological classification Euler's χ for topological classification ● 6 vertices = v ● 9 edges = e ● 4 faces = f ● χ = v – e + f ● χ = 1 for single components with no holes maverick.inria.fr Euler's χ for topological classification ● 16 vertices = v ● 28 edges = e ● 14 faces = f ● 2 “blobs” = b ● χ = v – e + f - b ● χ = (16+14)–(28+2) = 0 for “donuts” Euler's χ for topological classification ● 8 vertices = v ● 12 edges = e ● 6 faces = f ● 1 “blob” = b ● χ = v – e + f - b ● χ = 8 – 12 + 6 – 1 = 1 for “spheres” mathcraft.wonderhowto.com Euler's χ for topological classification ● 6+8 vertices = v ● (3+3/2)*8 edges = e ● 8*3 faces = f ● 1 “blob” = b ● χ = v – e + f - b ● χ = 14 - 36 + 24 – 1 = 1 for “spheres” mathcraft.wonderhowto.com Euler's χ for topological classification ● 20 + 12 vertices = v ● (3+3/2)*20 edges = e ● 20*3 faces = f ● 1 “blob” = b ● χ = v – e + f - b ● χ = 32 - 90 + 60 – 1 = 1 for “spheres” Angle deficit for topological classification Angle deficit for topological classification “And now for something completely different...” Angle deficit for topological classification ● Imagine a flattened cube ● At each vertex, 90 degrees is “missing” ● 8 * 90 = 720 ● Total Angle Deficit is 720 mathcraft.wonderhowto.com Angle deficit for topological classification ● A flattened cube! ● At each vertex, 90 degrees is “missing” ● 8 * 90 = 720 ● Total Angle Deficit is 720 Cuboctahedron Template Cuboctahedron in 3D ● 12 vertices ● 60 degree deficit per ● 12*60=720 ● Total Angle Deficit is 720 Sonobe! ● 20 vertices with 90 degree deficit (elliptic curvature!) ● 12 vertices with -90 degree deficit (hyperbolic curvature) ● 20*90-12*90=720 But.... Total Angle Deficit = 4π * χ Gauss-Bonnet Theorem Gauss-Bonnet Theorem If we trace out a closed curve on a surface, the total enclosed Gauss curvature (total angle deficit) is 2π minus the total angle defect (angles of deflection) around the curve. Gauss-Bonnet simple example ● Total angle defect is zero. (This paper ribbon is a geodesic.) ● Each half of the sphere must have the same enclosed Gauss curvature. ● Total must be 4π ● Each half contains 2π Gauss-Bonnet simple example ● Total angle defect is zero. (This paper ribbon is a geodesic.) ● Each half of the can must have the same enclosed Gauss curvature. ● Total must be 4π ● Each half contains 2π Gauss-Bonnet simple example ● Total angle defect is zero. ● Each half of the cube must have the same enclosed Gauss curvature. ● Each half contains 2π ● Each vertex has π/2 Gauss-Bonnet example ● Total angle defect is: 2(180-72)+2(180-120) 336 ● 360-336=24 degrees ● Angle deficit: 360-(60+60+108+108) 24 degrees Gauss-Bonnet example ● Total angle defect is: 3(180-84) 288 ● 360-288=72 degrees ● Angle deficit: 3(360-60-60-108-108) 720 degrees Gauss-Bonnet example ● Total angle defect is: 5(180-120) 300 ● 360-300 = 60 degrees ● Angle deficit: 5(360-108-120-120) 60 degrees Gauss-Bonnet example ● Total angle defect is zero (geodesic!) ● 360 degrees is promised ● Angle deficit: 5(90)+(-90) 360 degrees Bonus Puzzle Use 3 colors, 10 pieces of each color. Try to make sure you get exactly one of each color on each pyramid. (Solve using a graph first.) Bonus Puzzle Solution http://www.donnadietz.com http://www.donnadietz.com/Origami.pdf [email protected] .
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